Linear Algebra Examples

$[\begin{array}{c} 0 \\ 2 \\ 3 \end{array}]$ ,

$[\begin{array}{c} - 1 \\ - 3 \\ - 5 \end{array}]$ ,

$[\begin{array}{c} 2 \\ 0 \\ 1 \end{array}]$

Step 1

Assign the set to the name

$S$ to use throughout the problem.

$S = [\begin{array}{c} 0 \\ 2 \\ 3 \end{array}], [\begin{array}{c} - 1 \\ - 3 \\ - 5 \end{array}], [\begin{array}{c} 2 \\ 0 \\ 1 \end{array}]$

Step 2

Create a matrix whose rows are the vectors in the spanning set.

$[\begin{array}{c} 0 & 2 & 3 \\ - 1 & - 3 & - 5 \\ 2 & 0 & 1 \end{array}]$

Step 3

Find the reduced row echelon form of the matrix.

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Step 3.1

Swap

$R_{2}$ with

$R_{1}$ to put a nonzero entry at

$1, 1$ .

$[\begin{array}{c} - 1 & - 3 & - 5 \\ 0 & 2 & 3 \\ 2 & 0 & 1 \end{array}]$

Step 3.2

Multiply each element of

$R_{1}$ by

$- 1$ to make the entry at

$1, 1$ a

$1$ .

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Step 3.2.1

Multiply each element of

$R_{1}$ by

$- 1$ to make the entry at

$1, 1$ a

$1$ .

$[\begin{array}{c} - - 1 & - - 3 & - - 5 \\ 0 & 2 & 3 \\ 2 & 0 & 1 \end{array}]$

Step 3.2.2

Simplify

$R_{1}$ .

$[\begin{array}{c} 1 & 3 & 5 \\ 0 & 2 & 3 \\ 2 & 0 & 1 \end{array}]$

Step 3.3

Perform the row operation

$R_{3} = R_{3} - 2 R_{1}$ to make the entry at

$3, 1$ a

$0$ .

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Step 3.3.1

Perform the row operation

$R_{3} = R_{3} - 2 R_{1}$ to make the entry at

$3, 1$ a

$0$ .

$[\begin{array}{c} 1 & 3 & 5 \\ 0 & 2 & 3 \\ 2 - 2 \cdot 1 & 0 - 2 \cdot 3 & 1 - 2 \cdot 5 \end{array}]$

Step 3.3.2

Simplify

$R_{3}$ .

$[\begin{array}{c} 1 & 3 & 5 \\ 0 & 2 & 3 \\ 0 & - 6 & - 9 \end{array}]$

Step 3.4

Multiply each element of

$R_{2}$ by

$\frac{1}{2}$ to make the entry at

$2, 2$ a

$1$ .

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Step 3.4.1

Multiply each element of

$R_{2}$ by

$\frac{1}{2}$ to make the entry at

$2, 2$ a

$1$ .

$[\begin{array}{c} 1 & 3 & 5 \\ \frac{0}{2} & \frac{2}{2} & \frac{3}{2} \\ 0 & - 6 & - 9 \end{array}]$

Step 3.4.2

Simplify

$R_{2}$ .

$[\begin{array}{c} 1 & 3 & 5 \\ 0 & 1 & \frac{3}{2} \\ 0 & - 6 & - 9 \end{array}]$

Step 3.5

Perform the row operation

$R_{3} = R_{3} + 6 R_{2}$ to make the entry at

$3, 2$ a

$0$ .

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Step 3.5.1

Perform the row operation

$R_{3} = R_{3} + 6 R_{2}$ to make the entry at

$3, 2$ a

$0$ .

$[\begin{array}{c} 1 & 3 & 5 \\ 0 & 1 & \frac{3}{2} \\ 0 + 6 \cdot 0 & - 6 + 6 \cdot 1 & - 9 + 6 (\frac{3}{2}) \end{array}]$

Step 3.5.2

Simplify

$R_{3}$ .

$[\begin{array}{c} 1 & 3 & 5 \\ 0 & 1 & \frac{3}{2} \\ 0 & 0 & 0 \end{array}]$

Step 3.6

Perform the row operation

$R_{1} = R_{1} - 3 R_{2}$ to make the entry at

$1, 2$ a

$0$ .

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Step 3.6.1

Perform the row operation

$R_{1} = R_{1} - 3 R_{2}$ to make the entry at

$1, 2$ a

$0$ .

$[\begin{array}{c} 1 - 3 \cdot 0 & 3 - 3 \cdot 1 & 5 - 3 (\frac{3}{2}) \\ 0 & 1 & \frac{3}{2} \\ 0 & 0 & 0 \end{array}]$

Step 3.6.2

Simplify

$R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{1}{2} \\ 0 & 1 & \frac{3}{2} \\ 0 & 0 & 0 \end{array}]$

Step 4

Convert the nonzero rows to column vectors to form the basis.

${[\begin{array}{c} 1 \\ 0 \\ \frac{1}{2} \end{array}], [\begin{array}{c} 0 \\ 1 \\ \frac{3}{2} \end{array}]}$

Step 5

Since the basis has

$2$ vectors, the dimension of

$S$ is

$2$ .

$dim (S) = 2$

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